POSITION VECTORS & FORCE VECTORS
Transcript of POSITION VECTORS & FORCE VECTORS
POSITION VECTORS & FORCE VECTORS
Today’s Objectives:
Students will be able to :
a) Represent a position vector in Cartesian
coordinate form, from given geometry.
b) Represent a force vector directed along
a line.
In-Class Activities:
• Check Homework
• Reading Quiz
• Applications /
Relevance
• Write Position Vectors
• Write a Force Vector
• Concept Quiz
• Group Problem
• Attention Quiz
READING QUIZ
1. A position vector, rPQ, is obtained by
A) Coordinates of Q minus coordinates of P
B) Coordinates of P minus coordinates of Q
C) Coordinates of Q minus coordinates of the origin
D) Coordinates of the origin minus coordinates of P
2. A force of magnitude F, directed along a unit
vector U, is given by F = ______ .
A) F (U)
B) U / F
C) F / U
D) F + U
E) F – U
Wing strut
APPLICATIONS
How can we
represent the force
along the wing
strut in a 3-D
Cartesian vector
form?
POSITION VECTOR
A position vector is defined as
a fixed vector that locates a
point in space relative to
another point.
Consider two points, A & B, in
3-D space. Let their coordinates
be (XA, YA, ZA) and ( XB,
YB, ZB ), respectively.
The position vector directed from A to B, r AB , is defined as
r AB = {( XB – XA ) i + ( YB – YA ) j + ( ZB – ZA ) k }m
Please note that B is the ending point and A is the starting point.
So ALWAYS subtract the “tail” coordinates from the “tip”
coordinates!
FORCE VECTOR DIRECTED ALONG A LINE
(Section 2.8)
If a force is directed along a line,
then we can represent the force
vector in Cartesian Coordinates by
using a unit vector and the force
magnitude. So we need to:
a) Find the position vector, r AB , along two points on
that line.
b) Find the unit vector describing the line’s direction,
uAB = (rAB/rAB).
c) Multiply the unit vector by the magnitude of the
force, F = F uAB .
EXAMPLE
Given: 400 lb force along the
cable DA.
Find: The force FDA in the
Cartesian vector form.
EXAMPLE (continued)
The figure shows that when relating D
to A, we will have to go -2 ft in the x-
direction, -6 ft in the y-direction, and
+14 ft in the z-direction. Hence,
rDA = {-2 i – 6 j + 14 k} ft.
We can also find rDA by subtracting the
coordinates of D from the coordinates
of A.
rDA = (22 + 62 + 142)0.5 = 15.36 ft
uDA = rDA/rDA and FDA = 400 uDA lb
FDA = 400{(-2 i – 6 j + 14 k)/15.36} lb
= {-52.1 i – 156 j + 365 k} lb
CONCEPT QUIZ
1. P and Q are two points in a 3-D space. How are the position
vectors rPQ and rQP related?
A) rPQ = rQP B) rPQ = - rQP
C) rPQ = 1/rQP D) rPQ = 2 rQP
2. If F and r are force vector and position vectors, respectively,
in SI units, what are the units of the expression (r * (F / F)) ?
A) Newton B) Dimensionless
C) Meter D) Newton - Meter
E) The expression is algebraically illegal.
Given: Two forces are acting on a
pipe as shown in the figure.
Find: The magnitude and the
coordinate direction angles
of the resultant force.
GROUP PROBLEM SOLVING
GROUP PROBLEM SOLVING
(continued)
FCA = 100 lb{rCA/rCA}
FCA = 100 lb(-3 sin 40° i + 3 cos 40° j – 4 k)/5
FCA = {-38.57 i + 45.96 j – 80 k} lb
FCB = 81 lb{rCB/rCB}
FCB = 81 lb(4 i – 7 j – 4 k)/9
FCB = {36 i – 63 j – 36 k} lb
FR = FCA + FCB = {-2.57 i – 17.04 j – 116 k} lb
FR = (2.572 + 17.042 + 1162) = 117.3 lb = 117 lb
= cos-1(-2.57/117.3) = 91.3°, = cos-1(-17.04/117.3) = 98.4°
= cos-1(-116/117.3) = 172°
ATTENTION QUIZ
1. Two points in 3 – D space have coordinates of P (1, 2, 3) and Q
(4, 5, 6) meters. The position vector rQP is given by
A) {3 i + 3 j + 3 k} m
B) {- 3 i – 3 j – 3 k} m
C) {5 i + 7 j + 9 k} m
D) {- 3 i + 3 j + 3 k} m
E) {4 i + 5 j + 6 k} m
2. Force vector, F, directed along a line PQ is given by
A) (F/ F) rPQ B) rPQ/rPQ
C) F(rPQ/rPQ) D) F(rPQ/rPQ)
EXAMPLE
Given: 400 lb force along the
cable DA.
Find: The force FDA in the
Cartesian vector form.
Given: Two forces are acting on a
pipe as shown in the figure.
Find: The magnitude and the
coordinate direction angles
of the resultant force.
GROUP PROBLEM SOLVING